The positive integers x, y, and z are such that x is a factor of y and

y/x = k where k is an integer.
y = xk ....................i

z/y = m where m is an integer.
z = ym = xkm .....................ii

If a factor is even, then the source of the factor must be even.

1) If xz is even, z must be even because x may or may not be an even because x is a factor of z but z must be even. SUFF.
2) If y is even, z must be even because y is a factor of z. SUFF..

D..

Bunuel, you always try to solve the questions algebraically, don't you?

Not at all. There are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes). Also many questions can be solved with logic and common sense much quicker than with algebraic approach. So you shouldn't always rely on algebra. Having said that I must add that there are of course other types of questions which are perfect for algebraic approach, plus I often use algebra just to explain a solution.

I like to thing of the boxes method. If you draw them out, then x is inside y which is inside z.
zx, a 2 will exist inside the box of either z or x (which is itself inside z) so YES
y a 2 will exist inside the box of y which is itself z so YES

awesome. your explanations and bunuel as well, are amazing thanks

Easy if you realize the following:
When a is a factor of b
AND b is a factor of c
THEN a is a factor of c as well.

Hence when either one of these numbers is even, the other has to be even too..
D

z = ky

y = mx

so z = (km)xy

(1) -> xz is eve means at least x or z is even, and if x = even, then z is also even as it has an even factor.

2 -> y is even so z having an even factor is even too.

Answer D

At first, mistook "factor" for "multiple" came with answer E.
Later, understood that the problem was so easy...just plug and play !!

if we plug and play, why can't we test X = 1? then y/n..

The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even
2) y is even.
Please explain. thanks.[/quote]

Ans. let us take any 3 numbers, say x=3,y=18,z=54,
or x=2,y=4,z=20
1)if xz is even then it means that either x or z is even,say that x is even, now there is no even number which is a factor of odd number, so z is definitely even, now if x is odd as in the above case, still then we can point out that z is even.
2)if y is even then it is clear that z will be even.
Thus this question could be answered by any of the two questions.

Plug in method would be far more painful for this question (and most other questions in my opinion). Think how you would go about it:
Checking whether stmnt 1 is enough: xz is even
If x = 1, y = 1 and z = 2 (so that xz is even), then z is even.
If x = 2, y = 2 and z = 4, z is again even.
Then you start thinking if you can take some values such that xz is even but z is not... Now you start using logic... Wouldn't you say it is far better to use logic in the first place itself?

What is given?

\(z = y*N\)
\(y = x*R\)

1.
\(xz = 2*I\)
If x is even, then z is even since \(z = x*R\)
If z is even, then z is even.
SUFFICIENT!

2. \(y = 2*I\) ==> \(z = 2*I*N\) Definitely EVEN
SUFFICIENT!

Answer: D

Can you please provide a numerical example for this part it isn't very clear. Thanks in advance!

I thought it was insufficient as there were multiple cases either X or Y even or both....

We know that x is a factor of \(y \to y = Ix\)
Again, \(z = yI^'\). Now, given that xz - even.

Case I:Assume that x = even , z = odd. Now, as x is even, y = even(I can be odd/even,doesn't matter).
Again, as y is even, z HAS to be even(\(I^'\) is odd/even, doesn't matter). Thus, if x is even, z IS even.
Numerical Example :y = 2*I(x=2).
\(z = 6*I^'\). z IS even.

Case II : z is even OR (x and z) both are even.

Hope this helps.

Perfect explanation. Remember the factor foundation rule.
Also, other properties of factors that might be helpful to have in mind.

Just to remind you, The factor foundation rule states that "if a is a factor of b, and b is a factor of c, then a is a factor of c"
Also, if 'a' is a factor of 'b', and 'a' is a factor of 'c', then 'a' is a factor of (b+c). In fact, 'a' is a factor of (mb + nc) for all integers 'm' and 'n'
If 'a' is a factor of 'b' and 'b' is a factor of 'a', then 'a=b'
If 'a' is a factor of 'bc' and gcd (a,b) = 1, then 'a' is a factor of 'c'
If 'p' is a prime number and 'p' is a factor of 'ab' then 'p' is a factor of 'a' or 'p' is a factor of 'b'
In other words,, any integer is divisible by all of its factors- and it is also divisible by all of the factors of its factors

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Hope it helps

here is the question stem => y=ax z=by => z=cx for some integers a,b,c
now statement 1 => xz is even => cases are => E,E E,O O,E => If x is even then has to be even as z=cx and if x is odd => z has to be even as zx = even
=> suff
statement 2 => y is even => z=by => z has to be even as E*E = E
suff
SMASH that D

Bunuel if x, y, and z are 3, 9 and 27 repectivey then how can A be sufficient ?